Question

c. produces the greatest change in mass number? why?
d. produces the least change in mass number? why?

1. complete the following nuclear reactions:

a. $_{88}^{226}ra \
ightarrow _?^? ? + _{-1}^0 e$
b. $_{84}^{209}po \
ightarrow _{82}^{205}pb + _?^? ?$
c. $_{92}^{218}u \
ightarrow _?^? ? + _2^4 he$
d. $_{90}^{234}th \
ightarrow _{91}^{234}pa + _?^? ?$
e. $_?^? ? + _7^{14}n \
ightarrow _8^{17}o + _1^1 h$

1. when isotope bismuth-213 emits an alpha particle:

a. write out the nuclear equation:
b. which is the parent element?
c. which is the daughter element?
d. what new element results if the isotope, instead, emits a beta particle?

Explanation:

Step1: Analyze part a (Beta decay)

In beta decay (\(_{-1}^0e\) emission), the mass number remains the same, and the atomic number increases by 1. For \(_{88}^{226}\text{Ra}\), atomic number after decay: \(88 + 1=89\), mass number: \(226\). So the product is \(_{89}^{226}\text{Ac}\) (Actinium).
Equation: \(_{88}^{226}\text{Ra}
ightarrow _{89}^{226}\text{Ac} + _{-1}^0e\)

Step2: Analyze part b (Alpha-like? Wait, mass number change: \(209 - 205 = 4\), atomic number change: \(84 - 82 = 2\). So it's alpha particle (\(_{2}^4\text{He}\)) emission.

Check: Mass number: \(205 + 4 = 209\), atomic number: \(82 + 2 = 84\) (matches Po). So the particle is \(_{2}^4\text{He}\).
Equation: \(_{84}^{209}\text{Po}
ightarrow _{82}^{205}\text{Pb} + _{2}^4\text{He}\)

Step3: Analyze part c (Alpha decay, \(_{2}^4\text{He}\) emission)

Mass number of product: \(218 - 4 = 214\), atomic number: \(92 - 2 = 90\). So the product is \(_{90}^{214}\text{Th}\) (Thorium).
Equation: \(_{92}^{218}\text{U}
ightarrow _{90}^{214}\text{Th} + _{2}^4\text{He}\)

Step4: Analyze part d (Beta decay, \(_{-1}^0e\) emission? Wait, mass number same (\(234\)), atomic number: \(91 - 90 = 1\) decrease? Wait, no: \(_{90}^{234}\text{Th}

ightarrow _{91}^{234}\text{Pa}\), so the emitted particle has atomic number \(90 - 91=-1\), mass number \(234 - 234 = 0\). So it's \(_{-1}^0e\) (beta particle, but wait, Th to Pa: atomic number increases by 1, so Th emits \(_{-1}^0e\)? Wait, no: Th (90) to Pa (91): atomic number +1, so the emitted particle is \(_{-1}^0e\) (beta minus). Wait, the equation is \(_{90}^{234}\text{Th}
ightarrow _{91}^{234}\text{Pa} + _{-1}^0e\)? Wait, no, wait: the given equation is \(_{90}^{234}\text{Th}
ightarrow _{91}^{234}\text{Pa} + ?\). So mass number: \(234 = 234 + m\) → \(m = 0\). Atomic number: \(90 = 91 + z\) → \(z=-1\). So the particle is \(_{-1}^0e\) (beta particle).

Step5: Analyze part e (Reaction: \(? + _{7}^{14}\text{N}

ightarrow _{8}^{17}\text{O} + _{1}^{1}\text{H}\))
Let the unknown be \(_{z}^{A}X\). Mass number: \(A + 14 = 17 + 1\) → \(A = 4\). Atomic number: \(z + 7 = 8 + 1\) → \(z = 2\). So it's \(_{2}^4\text{He}\) (alpha particle).
Equation: \(_{2}^4\text{He} + _{7}^{14}\text{N}
ightarrow _{8}^{17}\text{O} + _{1}^{1}\text{H}\)

Answer:

s (for each subpart):
a. \(_{88}^{226}\text{Ra}
ightarrow \boldsymbol{_{89}^{226}\text{Ac}} + _{-1}^0e\)
b. \(_{84}^{209}\text{Po}
ightarrow _{82}^{205}\text{Pb} + \boldsymbol{_{2}^4\text{He}}\)
c. \(_{92}^{218}\text{U}
ightarrow \boldsymbol{_{90}^{214}\text{Th}} + _{2}^4\text{He}\)
d. \(_{90}^{234}\text{Th}
ightarrow _{91}^{234}\text{Pa} + \boldsymbol{_{-1}^0e}\)
e. \(\boldsymbol{_{2}^4\text{He}} + _{7}^{14}\text{N}
ightarrow _{8}^{17}\text{O} + _{1}^{1}\text{H}\)